Kth Largest XOR Coordinate Value

▶

Steps

setup

Initialize prefix XOR matrix and min-heap

Create a prefix XOR matrix with an extra row and column of zeros to simplify XOR calculations. Initialize an empty min-heap to store up to k XOR values.

💡 The prefix matrix with padding avoids boundary checks during XOR computations. The min-heap will track the top k XOR values efficiently.

Line:

m, n = len(matrix), len(matrix[0])
prefix = [[0] * (n + 1) for _ in range(m + 1)]
min_heap = []

💡 Prefix matrix and min-heap are prepared for the main computation.

fill_cells

Compute prefix[1][1] and insert into heap

Calculate prefix[1][1] = 0 ^ 0 ^ 0 ^ 5 = 5. Since heap size is less than k, insert 5 into the heap.

💡 The first XOR coordinate value is the matrix element at (0,0). Inserting it initializes the heap.

Line:

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]
val = prefix[i][j]
if len(min_heap) < k:
    heapq.heappush(min_heap, val)

💡 Heap grows until it contains k elements, ensuring candidates for kth largest value.

fill_cells

Compute prefix[1][2] and insert into heap

Calculate prefix[1][2] = 0 ^ 5 ^ 0 ^ 2 = 7. Heap size is less than k, so insert 7.

💡 This XOR value combines previous prefix XORs and the current matrix element to build cumulative XOR.

Line:

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]
val = prefix[i][j]
if len(min_heap) < k:
    heapq.heappush(min_heap, val)

💡 Heap now contains two XOR values, still building up to k elements.

heapify_up

Heapify up after inserting 7

Compare newly inserted 7 at index 1 with parent 5 at index 0. Since 7 > 5, no swap needed.

💡 Heapify up maintains min-heap property after insertion. Here, no swap occurs because 7 is greater than 5.

Line:

heapq.heappush(min_heap, val)

💡 Heap property is maintained without swaps when child is larger than parent in min-heap.

fill_cells

Compute prefix[2][1] and insert into heap

Calculate prefix[2][1] = 5 ^ 0 ^ 0 ^ 1 = 4. Heap size is less than k, insert 4.

💡 This XOR value represents the cumulative XOR for coordinate (2,1).

Line:

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]
val = prefix[i][j]
if len(min_heap) < k:
    heapq.heappush(min_heap, val)

💡 Heap now contains k=3 elements, ready for comparisons on next values.

heapify_up

Heapify up after inserting 4

Compare newly inserted 4 at index 2 with parent 4 at index 0. Since 4 == 4, no swap needed.

💡 Heapify up maintains min-heap property after insertion. Here, 4 is equal to parent, so heap property holds.

Line:

heapq.heappush(min_heap, val)

💡 Heap property is maintained when child is not less than parent in min-heap.

compare

Compute prefix[2][2] and compare with heap root

Calculate prefix[2][2] = 7 ^ 4 ^ 5 ^ 6 = 0. Compare 0 with heap root 4 to decide if replacement is needed.

💡 This XOR value is the cumulative XOR for coordinate (2,2). We compare it to the smallest in heap to decide if it should replace.

Line:

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]
val = prefix[i][j]
elif val > min_heap[0]:
    heapq.heapreplace(min_heap, val)

💡 Only values larger than the smallest in heap can improve the kth largest candidate.

traverse

All prefix XOR values processed; heap contains top k values

All prefix XOR values have been processed. The heap contains [4, 7, 5], representing the top 3 largest XOR coordinate values.

💡 The heap now holds the k largest XOR values, with the smallest among them at the root.

Line:

return min_heap[0]

💡 The root of the min-heap is the kth largest XOR coordinate value.

reconstruct

Return kth largest XOR coordinate value

Return the root of the min-heap, which is 4, the 3rd largest XOR coordinate value.

💡 The root of the min-heap after processing all values is the answer.

Line:

return min_heap[0]

💡 The algorithm efficiently returns the kth largest XOR value without full sorting.

final

Final heap state and answer

The min-heap contains the top k largest XOR coordinate values, with the root 4 being the kth largest.

💡 The final heap state confirms the answer and the heap structure after all operations.

Line:

return min_heap[0]

💡 Final heap state is consistent with the algorithm's output.

import heapq

def kthLargestValue(matrix, k):
    m, n = len(matrix), len(matrix[0])  # STEP 1
    prefix = [[0] * (n + 1) for _ in range(m + 1)]  # STEP 1
    min_heap = []  # STEP 1
    for i in range(1, m + 1):  # STEP 2-6
        for j in range(1, n + 1):  # STEP 2-6
            prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]  # STEP 2-6
            val = prefix[i][j]  # STEP 2-6
            if len(min_heap) < k:  # STEP 2-4
                heapq.heappush(min_heap, val)  # STEP 2-4
            elif val > min_heap[0]:  # STEP 5
                heapq.heapreplace(min_heap, val)  # STEP 5
    return min_heap[0]  # STEP 7

if __name__ == '__main__':
    matrix = [[5,2],[1,6]]
    k = 3
    print(kthLargestValue(matrix, k))  # Output: 4

📊

Kth Largest XOR Coordinate Value - Watch the Algorithm Execute, Step by Step

Watching each insertion and replacement in the heap reveals how the algorithm efficiently tracks the kth largest value without sorting all values explicitly.

insertmin-heap

empty heap

insertmin-heap

5

0

insertmin-heapswap [0↔1]

5

0

7

1

heapify_upmin-heap

5

0

7

1

insertmin-heapswap [0↔2]

5

0

7

1

4

2

heapify_upmin-heap

4

0

7

1

5

2

peekmin-heap

4

0

7

1

5

2

peekmin-heap

4

0

7

1

5

2

Added: 4

peekmin-heap

4

0

7

1

5

2

Added: 4

peekmin-heap

4

0

7

1

5

2

Added: 4

Key Takeaways

✓ The min-heap efficiently maintains the top k largest XOR values without sorting all values.

This insight is hard to see from code alone because the heap operations are implicit; visualization shows the heap content changing step-by-step.

✓ Prefix XOR matrix allows constant-time calculation of XOR coordinate values by reusing previous computations.

Understanding prefix XOR is easier when you see how each cell depends on neighbors, rather than just reading the formula.

✓ Only XOR values larger than the heap root replace the smallest in the heap, pruning smaller candidates dynamically.

The decision to replace or not is clearer when you see the heap root compared to the current value visually.

Practice

1. Given the following Python code implementing the optimal approach for Kth Largest in a Stream, what is the output of the last print statement?

import heapq

class KthLargest:
    def __init__(self, k: int, nums: list[int]):
        self.k = k
        self.heap = nums
        heapq.heapify(self.heap)
        while len(self.heap) > k:
            heapq.heappop(self.heap)

    def add(self, val: int) -> int:
        if len(self.heap) < self.k:
            heapq.heappush(self.heap, val)
        elif val > self.heap[0]:
            heapq.heapreplace(self.heap, val)
        return self.heap[0]

kthLargest = KthLargest(3, [4, 5, 8, 2])
print(kthLargest.add(3))
print(kthLargest.add(5))
print(kthLargest.add(10))
print(kthLargest.add(9))
print(kthLargest.add(4))

easy

A. 8

B. 5

C. 4

D. 9

Solution

1. Step 1: Trace heap initialization

   Initial nums = [4,5,8,2], heapified and trimmed to size k=3: heap becomes [4,5,8] after popping 2.

2. Step 2: Trace add calls and heap updates

   add(3): 3 < 4 (heap[0]), no replace, return 4 add(5): 5 > 4, replace 4 with 5, heap now [5,5,8], return 5 add(10): 10 > 5, replace 5 with 10, heap now [5,10,8], return 5 add(9): 9 > 5, replace 5 with 9, heap now [8,10,9], return 8 add(4): 4 < 8, no replace, return 8

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Last add returns 8 as kth largest element [OK]

Hint: Heap root always holds kth largest element [OK]

Common Mistakes:

• Off-by-one errors in heap size
• Replacing when val <= heap[0]
• Misunderstanding heap root meaning

2. You need to hire exactly K workers from a pool, each with a quality and a minimum wage expectation. You must pay every hired worker at least their minimum wage, and all hired workers must be paid in the same ratio to their quality. Which approach guarantees finding the minimum total cost to hire exactly K workers under these constraints?

easy

A. Use a brute force approach checking all subsets of size K and calculating total cost for each.

B. Use dynamic programming to select workers minimizing total wage while respecting quality constraints.

C. Sort workers by quality and greedily pick the first K workers with the lowest quality.

D. Sort workers by their wage-to-quality ratio, then use a max-heap or balanced tree to maintain the K smallest qualities while iterating through workers in ratio order.

Solution

1. Step 1: Understand the fairness constraint

   All hired workers must be paid at the same wage-to-quality ratio, so the maximum ratio among chosen workers determines the pay rate.

2. Step 2: Use ratio sorting and a data structure to maintain K workers

   Sorting by ratio ensures we consider possible pay rates in ascending order. Maintaining the K smallest qualities with a max-heap or balanced tree minimizes total quality sum, thus minimizing total cost = ratio * sum of qualities.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Sorting by ratio + max-heap for qualities is the known optimal approach [OK]

Hint: Sort by ratio and maintain K smallest qualities [OK]

Common Mistakes:

• Greedy by quality only ignores wage constraints
• DP approach is not efficient or straightforward here
• Brute force is correct but not efficient

3. The following code attempts to find the maximum CPU load from a list of jobs. Identify the line containing the subtle bug that causes incorrect results in some cases.

def find_max_cpu_load(jobs):
    events = []
    for start, end, load in jobs:
        events.append((start, load))
        events.append((end, load))  # Bug here
    events.sort(key=lambda x: x[0])

    current_load = 0
    max_load = 0
    for _, load_change in events:
        current_load += load_change
        max_load = max(max_load, current_load)
    return max_load

medium

A. Line adding (end, load) instead of (end, -load) to events.

B. Line sorting events by start time only.

C. Line initializing current_load to 0 before loop.

D. Line updating max_load inside the loop.

Solution

1. Step 1: Identify event creation lines.

   Start events add positive load, end events must subtract load.

2. Step 2: Spot that (end, load) incorrectly adds load instead of subtracting.

   This causes current_load to accumulate incorrectly, inflating max_load.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   End event load must be negative to reduce current load [OK]

Hint: End events must subtract load, not add [OK]

Common Mistakes:

• Forgetting negative load at job end
• Misordering events with same timestamp

4. Consider the following buggy code snippet for the minimum cost to hire K workers problem. Which line contains the subtle bug that causes incorrect results on some inputs?

import bisect

def mincostToHireWorkers(quality, wage, K):
    workers = sorted([(w/q, q) for q, w in zip(quality, wage)])
    qualities = []
    sum_quality = 0
    min_cost = float('inf')
    for ratio, q in workers:
        bisect.insort(qualities, q)
        sum_quality += q
        if len(qualities) > K:
            removed = qualities.pop(0)  # BUG: removes smallest quality instead of largest
            sum_quality -= removed
        if len(qualities) == K:
            cost = ratio * sum_quality
            if cost < min_cost:
                min_cost = cost
    return min_cost

medium

A. Line where qualities.pop(0) removes the smallest quality

B. Line where qualities are inserted using bisect.insort

C. Line where workers are sorted by ratio

D. Line where cost is calculated as ratio * sum_quality

Solution

1. Step 1: Identify the data structure usage

   qualities is a sorted list; bisect.insort inserts maintaining ascending order.

2. Step 2: Check removal logic when size exceeds K

   qualities.pop(0) removes the smallest quality, but we want to remove the largest quality to minimize sum_quality.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Removing smallest quality increases sum_quality, leading to incorrect cost [OK]

Hint: Remove largest quality, not smallest, to minimize sum [OK]

Common Mistakes:

• Removing smallest quality instead of largest
• Not updating min_cost when size == K
• Sorting workers incorrectly

5. Suppose you want to find the top k frequent words but now words can be reused infinitely many times (like unlimited supply). Which modification to the optimal bucket sort approach correctly handles this scenario?

hard

A. This problem reduces to a knapsack variant; bucket sort no longer applies

B. Modify frequency counting to consider infinite reuse, then apply bucket sort as usual

C. Use a dynamic programming approach to count combinations of words for frequencies

D. Use a max heap keyed by frequency and word lex order, repeatedly pop top k elements

Solution

1. Step 1: Understand infinite reuse impact

   Infinite reuse means frequency counts are not fixed; problem becomes combinatorial.

2. Step 2: Recognize problem class change

   Top k frequent words with unlimited reuse resembles knapsack or coin change, not simple frequency counting.

3. Step 3: Bucket sort no longer applies

   Bucket sort relies on fixed frequencies; dynamic programming or other combinatorial methods are needed.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Infinite reuse breaks frequency counting assumptions [OK]

Hint: Infinite reuse changes problem class, bucket sort invalid [OK]

Common Mistakes:

• Trying to reuse bucket sort with infinite counts
• Ignoring combinatorial explosion
• Assuming heap approach suffices